What are the absolute extrema of #y=cos^2 x - sin^2 x# on the interval [-2,2]? Calculus Graphing with the First Derivative Identifying Turning Points (Local Extrema) for a Function 1 Answer Jim H Apr 20, 2015 #cos^2x-sin^2x = cos(2x)# which has a maximum value of #1# (at #x=0#) and a minimum value of #-1# (at #2x = pi# so #x = pi/2#) Answer link Related questions How do you find the x coordinates of the turning points of the function? How do you find the turning points of a cubic function? How many turning points can a cubic function have? How do you find the coordinates of the local extrema of the function? How do you find the local extrema of a function? How many local extrema can a cubic function have? How do I find the maximum and minimum values of the function #f(x) = x - 2 sin (x)# on the... If #f(x)=(x^2+36)/(2x), 1 <=x<=12#, at what point is f(x) at a minimum? How do you find the maximum of #f(x) = 2sin(x^2)#? How do you find a local minimum of a graph using the first derivative? See all questions in Identifying Turning Points (Local Extrema) for a Function Impact of this question 2250 views around the world You can reuse this answer Creative Commons License